Malfatti-Steiner Problem I. A. Sakmar, University of ... - MAA Sections

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We just proved that the opposed angles at C 2 and D in the quadrilateral AC 2 FD add up to 180 o . Now in the quadrilateral MC 2 FD the tangent C 2 F is ⊥ to the radius MC 2 . Thus the angle ∠ MC 2 F = 90 o . Similarly FD ⊥ MD. Thus the angle ∠ FDM = 90 o . Hence the sum of these two opposite angles is 180 o . Also, the two quadrilaterals have the angle ∠ C 2 FD common. Thus ∠ C 2 MD = ∠ C 2 AD. Now connect A to F. We will prove that AF is the bisector of the angle ∠ C 2 MD=∠ A. Let D’ be the point on AB with FC 2 = FD’. The triangles AD’F = ADF. Because: 1) AF = AF is common to both triangles. 2) FD = FD’ .Because FD=FC 2 are tangents to the circle S M and we took FD’ = FC 2 . 3) ∠ AD’F = ∠ ADF. Because the triangle C 2 FD’ being isosceles ∠ ADF is complementary to ∠ AC 2 F. But we also showed that ∠ FDA is complementary to ∠ AC 2 F. Hence AD’F = ADF Fig.5 We now claim that the points a, ß, C 1 , and C 2 lie on a circle . We shall call this circle S t and prove the claim below. -6-
Fig.6 Theorem: Given two mutually tangent circles S a , and S b which touch at ?. Consider the circles S c , and S d which touch S a , and S b ats the points a, ß, d, and e. The points a, ß, d, and e lie on a circle. The proof is again by inversion. Take ? as the center of inversion. S a , and S b become under inversion two lines l a , and l b . the point ? goes to infinity. Since S a , and S b do not have a second intersection point (they are tangent) they become parallel lines under inversion. S c , and S d are tangent to both S a , and S b . therefore their images must be tangent to the inverted images of S a , and S b . Fig.7 Obviously the points a’, ß’, d’, and e’ lie on a circle. Because they lie at the corners of a rectangle. Hence the inverted image of this circle goes through the points a, ß, d, and e. In our case S d is degenerate ( it is the line C 1 C 2 tangent to the circles Sa, and S b ). Thus we established that the points a, ß, C 1 , and C 2 lie on a circle , we called S t . -7-
Page 1 and 2: Malfatti-Steiner Problem I.A.Sakmar
Page 3 and 4: Call the intersection of S M with t
Page 5: In our case the circles S 1 , and S
Page 9 and 10: This shows that E is on the circle
Page 11 and 12: The circle S C which was tangent to
Page 13 and 14: This means that E is on the bisecto
Page 15 and 16: Fig.14 Since Steiner’s line is so
Page 17 and 18: A C K N O W L E D G M E N T S I ded

Malfatti-Steiner Problem I. A. Sakmar, University of ... - MAA Sections

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